A research study examined the blood vitamin D levels of the entire US population of landscape gardeners. The population average level of vitamin D in US landscapers was found to be 45.1 ng/mL with a standard deviation of 5.531 ng/mL. Assuming the true distribution of blood vitamin D levels follows a normal distribution, if you randomly select a landscaper in the US, what is the likelihood that his/her vitamin D level will be 54.75 ng/mL or more

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step-by-step explanation:

345

step-by-step explanation:

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answer:   (7x + 6y) ( 49x^2 - 42xy + 36y^2)

discussion:

in general, the factorization of a^3 + b^3 is

a^3 + b^ 3 = (a + b) ( a^2 - ab + b^2)

in our case note that 343 = 7^3 and 216 = 6^3 so the equation becomes:

343x^3 + 216y^3 = (7x)^3 + (6y)^3

if you think of a (above) as 7x and b as 6y the factorization becomes

343x^3 + 216y^3 =

((7x + 6y) (   (7x)^2 - 7x*6y + (6y)^2)   =

(7x + 6y) ( 49x^2 - 42xy + 36y^2)

you,

mrb

(a + b) (a^2 - a b + b^2)